2  Chapter 1: Discrete Probability Distributions

2.1 Module 1: Discrete Probability Distributions

2.1.1 Topic Summary

This module covers the fundamentals of discrete random variables and their probability distributions. A Probability Mass Function (PMF) assigns probabilities to each possible outcome, ensuring the sum equals 1. Key concepts include expected value \(E[X] = \sum x p(x)\) and variance \(Var(X) = E[X^2] - (E[X])^2\). The Binomial distribution models the number of successes in n trials with success probability p, with mean \(np\) and variance \(np(1-p)\), used for problems involving fair coins or defective items. The Poisson distribution approximates Binomial for rare events, with mean and variance both \(\lambda\), applied to arrival counts or defect probabilities. Joint PMF for two variables allows finding marginal distributions by summing over the other variable, essential for understanding dependencies between discrete outcomes (Ross 2019).

Tips for Beginners
  • PMF Basics: Remember, for any PMF, the probabilities must add up to 1. Always check this first in problems.
  • Mean and Variance: Mean is the “average” value, variance measures spread. For calculations, list all \(x\) and \(p(x)\), multiply, and sum. For variance, first find \(E[X^2]\) by squaring \(x\) before multiplying.
  • Binomial vs Poisson: Use Binomial for fixed trials (e.g., coin flips). Poisson for rare events over time/space (e.g., defects per hour). Poisson approximates Binomial when \(n \geq 20\) and \(p \leq 0.1\).
  • Common Mistakes: Don’t forget factorials in Poisson! For joint PMF, marginal is sum over the other variable – think of it as “collapsing” the table.
  • Practice: Start with simple cases, like \(n=2\) for Binomial, to build intuition before tackling larger problems.

2.1.2 Solved Problems

  1. Poisson Probability:

If \(X\) is a Poisson random variable such that \(P(X=0) = P(X=1)\), find \(P(X=2)\).

  • Solution: Using the formula \(P(X=x) = \frac{e^{-\lambda}\lambda^x}{x!}\), equating \(P(0) = P(1)\) yields \(\frac{e^{-\lambda}\lambda^0}{0!} = \frac{e^{-\lambda}\lambda^1}{1!}\). This simplifies to \(1 = \lambda\). Substituting \(\lambda=1\) to find \(P(2) = \frac{e^{-1}(1)^2}{2!} = \frac{1}{2e} \approx 0.1839\).

  1. Binomial Parameters: The variance of a Binomial distribution with \(n=10\) is 2.4. Find the mean of the distribution.

    • Solution: Variance is \(npq = 2.4\), which means \(10p(1-p) = 2.4\). Solving the quadratic equation \(p^2 - p + 0.24 = 0\) gives \(p=0.4\) or \(p=0.6\). Assuming \(p=0.4\), the Mean \(= np = 10 \times 0.4 = 4\).

  2. PMF Properties: A discrete random variable has PMF \(p(x) = x/10\) for \(x=1,2,3,4\). Find \(P(X < 3)\), \(E[X]\), and \(Var(X)\).

    • Solution: \(P(X < 3) = P(1) + P(2) = 0.1 + 0.2 = 0.3\). \(E[X] = \sum x P(x) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 3\). \(Var(X) = E[X^2] - (E[X])^2 = 10 - 3^2 = 10 - 9 = 1\).

  3. Binomial Application: 10% of items produced by a machine are defective. Find the probability that in a sample of 10 items, exactly 2 are defective.

    • Solution: This is Binomial with \(n=10\) and \(p=0.1\). \(P(X=2) = 10C_2 (0.1)^2 (0.9)^8 \approx 0.1937\).

  4. Joint PMF: The joint PMF of \((X,Y)\) is \(p(x,y) = k(x+y)\) for \(x=1,2\) and \(y=1,2\). Find the value of \(k\) and the marginal distribution of X.

    • Solution: The sum of all probabilities must equal 1: \(k[(1+1) + (1+2) + (2+1) + (2+2)] = 1 \implies 12k = 1 \implies k = 1/12\). The marginal for \(X\) is the sum over \(y\): \(P(X=1) = \sum_{y=1}^2 \frac{1+y}{12} = \frac{2+3}{12} = \frac{5}{12}\).

  5. Binomial Mean: A fair coin is tossed 3 times. Let \(X\) be the number of heads. Find the mean of \(X\).

    • Solution: This is a binomial distribution with \(n=3\) and \(p=0.5\). The mean is \(E[X] = np = 3 \times 0.5 = 1.5\).

  6. Poisson Probability: If \(X\) is a Poisson random variable such that \(P(X=1) = P(X=2)\), find \(P(X=0)\).

    • Solution: Equating the terms: \(\frac{e^{-\lambda}\lambda^1}{1!} = \frac{e^{-\lambda}\lambda^2}{2!}\), which simplifies to \(\lambda = 2\). Thus, \(P(X=0) = e^{-2}\).

  7. Joint PMF Marginals: The joint PMF of \((X, Y)\) is given by \(P(x, y) = \frac{x+y}{21}\) for \(x = 1, 2\) and \(y = 1, 2, 3\). Find the marginal distributions of \(X\) and \(Y\).

    • Solution: By summing across rows/columns, Marginal \(X\) is \(P(X=1) = 9/21, P(X=2) = 12/21\). Marginal \(Y\) is \(P(Y=1) = 5/21, P(Y=2) = 7/21, P(Y=3) = 9/21\).

2.1.3 Practice Problems

  • Suppose \(X\) is a binomial random variable with parameters \(n = 100\) and \(p = 0.02\). Find \(P(X < 3)\) using Poisson approximation to \(X\).

  • Derive the mean and variance of a Poisson distribution with parameter \(\lambda\).

  • Derive the mean and variance of a Binomial distribution.

  • A discrete random variable X has the following probability distribution: \(X : 0, 1, 2, 3\) and \(P(X=x) : 0.1, k, 2k, 0.3\). Find (i) the value of \(k\), (ii) \(P(X > 1)\), and (iii) \(E[X]\).

  • In a large consignment of electric bulbs, 5% are defective. A sample of 100 bulbs is taken. Use Poisson approximation to find the probability that at most 2 bulbs are defective.

Exam Preparation Tips
  • Memorize formulas: Binomial P(X=k) = C(n,k) p^k (1-p)^{n-k}, Poisson P(X=k) = e^{-λ} λ^k / k!.
  • For approximations, check conditions: n large, p small for Poisson.
  • Practice calculating combinations (nCr) manually for small n, use calculator for large.
  • When stuck, draw tables for PMF or joint distributions to visualize.